Question

In an Otto cycle, air is compressed adiabatically from ${27}^{o}C$ and 1 bar to 12 bar. heat is supplied at constant volume until the pressure rises to 35 bar. For the air $\gamma =1.4,c-v=0.718kJ/kgK$ and R=0.2872 kJ/kgK. What is mean effective pressure of the cycle?

• A. $596.502$
• B. $4.9$ bar
• C. $9.07$ bar
• D. $5.96$ bar

Answer 1

The correct option is D $5.96$ bar

$T_1=27+273$

$T_1=300K$

$\frac{T_2}{T_1}={\left(\frac{P_2}{p_1}\right)}^{(\gamma -1/\gamma )}$

$\frac{T_2}{T_1}=(12{)}^{(1.6-1)/1.4}$

$T_2=610.181K$

as (2-3) constant volume process

$T_4=T_3\times$

$\frac{P_2}{P_3}=\frac{T_2}{T_3}$

$T_3=\frac{P_3}{P_2}\times T_2=\frac{35}{12}\times 610.181$

$T_3=1779.694K$

$r=\frac{V_1}{V_2}={\left(\frac{P_2}{p_1}\right)}^{1/\gamma }$

r=5.899

For 3-4 process

$T_5{V}_3^{\gamma -3}=T_4{V}_4^{\gamma -1}$



$T_4=T_3\times {\left(\frac{V_3}{V_4}\right)}^{\gamma -1}$

$T_4=T_3\times {\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$

$T_4=\frac{T_3}{(5.894{)}^{\gamma -1}}$

$T_4=875.0522K$

$W_C=c_2(T_2-T_1)=222.71kJ/kg$

$W_T=c_2(T_3-T_4)=649.5328kJ/kg$

$W_{net}=W_T-W_C=426.822kJ/kg$

$W_{net}=mep\times V_S$

$V_S=(V_1-V_2)=V_1)(1-\frac{V_2}{V_1})=\frac{R{T}_1}{P_1}(1-\frac{1}{5.899})$

$=\frac{0.2872\times 300}{100}\times 0.83048$

$V_5=0.7155m^5/kg$

So, mep $=\frac{W_{net}}{V_S}=1-\frac{1}{(r{)}^{\gamma -1}}=1-\frac{1}{(5.89{)}^{1.4-1}}=0.508$

$Q=c_p(T_3-T_2)=0.718\times (1779.69-610.18)=839.71kJ/kg$

$W_{net}=Q\times {\eta }_{otto}=839.71\times 0.508=426.58kJ/kg$

$P_{mep}\times v_1(1-\frac{1}{r})$

$426.58=P_{mep}\times \frac{0.287\times 300}{100}(1-\frac{1}{5.89})$

$P_{mep}=586.77kPa=5.96$ bar